Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\left\\{\begin{array}{ll} \frac{1-\cos \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0) \end{array}\right.$$

Short answer

Based on the analysis and solution, determine the points in ℝ² where the given function is continuous. Answer: The function is continuous for all points in ℝ², including (0, 0).

Step-by-step solution

Continuity at \((x, y) \neq (0, 0)\)

For all points except \((0, 0)\), the function is given by \(f(x, y) = \frac{1-\cos\left(x^2 + y^2\right)}{x^2 + y^2}\). Since both the numerator and the denominator are continuous functions, their quotient is continuous wherever the denominator is nonzero. As \(x^2 + y^2\) is never zero when \((x, y) \neq (0, 0)\), the function is continuous for all \((x, y) \neq (0, 0)\).

Continuity at \((x, y) = (0, 0)\)

To analyze the continuity of the function at the origin, we will first switch to polar coordinates: \(x = r \cos\theta\) and \(y = r\sin\theta\). The function \(f(x, y)\) then becomes: $$f(r, \theta) = \frac{1-\cos\left(r^2\right)}{r^2}$$ We want to find the limit of \(f(r, \theta)\) as \(r\) goes to 0 (irrespective of \(\theta\)) to check the continuity at \((0, 0)\). Note that we have $$0 \leq 1 - \cos(r^2) \leq 1 - \cos(0) = 0$$ Divide each term by \(r^2\): $$0 \leq \frac{1 - \cos(r^2)}{r^2} \leq \frac{0}{r^2}=0$$ Applying the squeeze theorem, we conclude that the limit exists and is equal to 0. Hence, \(f(x, y)\) is continuous at \((0, 0)\).

Conclusion

The function is continuous for all points in \(\mathbb{R}^2\) including \((0, 0)\).

Key concepts

Limits and Continuity

Understanding the continuity of functions in multivariable calculus starts with grasping the core concept of limits. Simply put, a function of two variables is continuous at a point if it smoothly passes through that point without any jumps or abrupt changes in value.

In the exercise, we examined continuity for the function at all points in \(\mathbb{R}^2\). A key aspect of establishing continuity at a point \( (x, y) \eq (0, 0) \) is showing that the limit of the function as \( (x, y) \) approaches that point equals the function's value at the point. Since both the numerator and denominator in our function are continuous away from the origin and the denominator doesn't vanish, continuity is guaranteed at those points. The crux is checking the point where the denominator could be zero, which, in this case, is at the origin, \( (0, 0) \).

Squeeze Theorem

The squeeze theorem is a handy tool for determining limits, particularly when it's tough to compute them directly. Imagine 'squeezing' a function between two others to find out its limit. If the upper and lower bounding functions converge to the same value at a certain point, the squeezed function must also converge to that same value.

In our exercise, we applied the squeeze theorem by bounding the function \( f(r, \theta) \) between zero and itself, which means that as \( r \) approaches zero, the squeeze theorem confirms that \( f(r, \theta) \) approaches zero as well. This elegant method sidesteps complicated limit calculations and proves the continuity at the origin unambiguously.

Polar Coordinates

Conversion to Polar Coordinates

When dealing with multivariable functions, particularly those that involve circles or spheres, it's often useful to switch from Cartesian to polar coordinates. This transformation simplifies the functions and, consequently, the calculations of limits and continuity. The conversion is given by \( x = r \cos\theta \) and \( y = r \sin\theta \), where \( r \) is the distance from the origin, and \( \theta \) is the angle from the positive x-axis.

In the context of the given exercise, converting to polar coordinates made it much easier to analyze the behavior of the function near the origin by reducing the two independent variables \( x \) and \( y \) to one, \( r \) (with \( \theta \) not influencing the limit as \( r \) approaches zero). This technique is particularly useful when dealing with functions that exhibit circular symmetry.

Multivariable Functions

Multivariable functions, like the one in our exercise, are functions that take several inputs and provide one output. These functions are often more complex to analyze, especially when it comes to concepts like continuity and limits. Unlike single-variable calculus, where you only consider functions along a line, here, you need to think in higher dimensions.

To determine the continuity of a multivariable function, one usually checks that the limit exists and is equal to the function's value at that point for all paths leading to the point in question. If a function passes this test for every point in its domain, as our function \( f(x, y) \) did, then it is continuous throughout the domain.